int 1 / (sqrt x + ^3sqrt x ) dx (hint: let u = 6 sqrt x, and

[crzymath]

need help with this problem:

dont know how to start off

int = integration

int 1 / (sqrt x + [sup:2uz0cm57]3[/sup:2uz0cm57]sqrt x ) dx

[Hint: Let u = 6 sqrt x, and write sqrt x and [sup:2uz0cm57]3[/sup:2uz0cm57]sqrt x in terms of u.]

 

[galactus]

Re: tough integration problem

This this what you mean?.:

$\displaystyle \int\frac{1}{\sqrt{x}+\sqrt[3]{x}}dx$


If we make the sub $\displaystyle u=x^{\frac{1}{6}}, \;\ u^{6}=x, \;\ u^{5}du=\frac{1}{6}dx$

Making the subs gives us:

$\displaystyle \int\left(6u^{2}-6u+6-\frac{6}{u+1}\right)du$

 

[crzymath]

Re: tough integration problem

thanks-that seemed easier than i thought it would be

by the way-how were you able to type in the equation??